First slide

Equilibrium of a particle

Question

P, Q and R are three coplanar forces acting at a point and are in equilibrium. Given P=1.9318 kg wt., $\sin θ_{1} = 0.9659$ , the value of R is ( in kg wt.)

Moderate

A

0.9659
B

2
C

1
D

$\frac{1}{2}$

Solution

$\frac{P}{\sin θ_{1}} = \frac{Q}{\sin θ_{2}} = \frac{R}{\sin 150^{\circ}}$
$\begin{array}{l} \Rightarrow \frac{1.93}{\sin θ_{1}} = \frac{R}{\sin 150^{\circ}} \\ \Rightarrow R = \frac{1.93 \times \sin 150^{\circ}}{\sin θ_{1}} = \frac{1.93 \times 0.5}{0.9659} = 1 \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App